I’m converting polar equations to rectangular form and I also want to, in the end, identify the resulting curve. I’m starting out with r equals 4 over 1 plus sine theta.

Now my instinct here is to multiply the 1 plus sine theta out of the denominator, because it will give me on the left side an r sine theta. That’s something I can work with. So let me do that.

I’m going to multiply both sides by 1 plus sine theta. And I get r times 1 plus sine theta equals just 4. So I have r, plus r sine theta equals 4. Now the r sine theta is y, that’s a small victory here, but I still have this r to deal with. So what I want to do is I’m going to isolate the r. I’ll tell you why in a second. So 4 minus y.

Let’s go back to our formulas here. I know that r² is x² plus y². Whenever I have an isolated r, I can square both sides to get an r² and then that will become an x² plus y². So let me do that. I'll square both sides and here I get 16 minus 8 y plus y², and the r² becomes x² plus y². And here I’ll get a little cancellation which is nice the y²'s, are going to go away. I’m left with x² equals 16 minus 8y.

Let me switch places, I’ll bring the 8y over and 16 minus x² and then I’ll divide both sides by 8. I get y equals 2 minus 1/8 x². This is a parabola opening downward with vertex at 0,2.

Parabola, vertex 0,2 opens down. And so it’s kind of interesting that you can get the equation of a parabola in polar. This is what it looks like. We’ll be dealing with more with the conic sections in polar form later on.